Solve:
\[6 = 3^x \]
Tolerance is \(\pm0.01\).
Solve:
\[10 = 9^x \]
Tolerance is \(\pm0.01\).
Solve:
\[20 = 2^x \]
Tolerance is \(\pm0.01\).
Solve:
\[26 = 2^x \]
Tolerance is \(\pm0.01\).
Solve:
\[2 = 6^x \]
Tolerance is \(\pm0.01\).
Solve:
\[4 = 9^x \]
Tolerance is \(\pm0.01\).
Solve:
\[18 = 2^x \]
Tolerance is \(\pm0.01\).
Solve:
\[15 = 9^x \]
Tolerance is \(\pm0.01\).
Solve:
\[29 = 4^x \]
Tolerance is \(\pm0.01\).
Solve:
\[34 = 9^x \]
Tolerance is \(\pm0.01\).
Let’s say a warm block initially at 74.5°C is placed outside on a moderately windy day. The outdoor temperature is 14.9°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.08 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.08(T-14.9)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 8.664 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 59.6e^{-0.08 t}+14.9\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 22.1 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 62.9°C is placed outside on a moderately windy day. The outdoor temperature is 15°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.08 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.08(T-15)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 8.664 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 47.9e^{-0.08 t}+15\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 29.2 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 84.5°C is placed outside on a moderately windy day. The outdoor temperature is 12.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-12.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 72.3e^{-0.32 t}+12.2\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 27.6 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 58.1°C is placed outside on a moderately windy day. The outdoor temperature is 24.4°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.07 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.07(T-24.4)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 9.902 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 33.7e^{-0.07 t}+24.4\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 35 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 97.5°C is placed outside on a moderately windy day. The outdoor temperature is 5.6°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.24 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.24(T-5.6)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.888 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 91.9e^{-0.24 t}+5.6\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 14.8 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 100.7°C is placed outside on a moderately windy day. The outdoor temperature is 25.6°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.06 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.06(T-25.6)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 11.552 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 75.1e^{-0.06 t}+25.6\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 37.9 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 96.7°C is placed outside on a moderately windy day. The outdoor temperature is 26.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-26.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 70.5e^{-0.32 t}+26.2\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 58.7 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 55.5°C is placed outside on a moderately windy day. The outdoor temperature is 4.8°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.32 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.32(T-4.8)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 2.166 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 50.7e^{-0.32 t}+4.8\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 18.8 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 79.8°C is placed outside on a moderately windy day. The outdoor temperature is 18.2°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.09 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.09(T-18.2)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 7.702 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 61.6e^{-0.09 t}+18.2\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 48.2 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s say a warm block initially at 103.9°C is placed outside on a moderately windy day. The outdoor temperature is 17°C. Moment by moment, the instantaneous rate of cooling is proportional to the difference of temperatures. The exact rate depends of many components:
For this block, all the effects combine into a single proportionality constant: 0.04 \(\text{min}^{-1}\). So, at any moment, the rate of change equals \(-0.04(T-17)\) \(\frac{^\circ\text{C}}{\text{min}}\). This implies that every 17.329 minutes, the temperature difference halves.
By reviewing Newton’s law of cooling, we see this block’s temperature follows an offset-exponential function of time.
\[T(t) ~=~ 86.9e^{-0.04 t}+17\]
where \(T\) is in degrees Celsius and \(t\) is in minutes. How many minutes does it take for the block’s temperature to reach 37.7 degrees Celsius?
The tolerance is \(\pm0.01\) min.
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 22°C.
\[T(0) = 22 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.44 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 13°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.44 T+13\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.44T+13)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 22 | 0.1 | 0.332 |
| 0.1 | 22.332 | 0.1 | 0.317392 |
| 0.2 | 22.649392 | 0.1 | 0.3034268 |
| 0.3 | 22.9528188 | 0.1 | 0.290076 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 22+0.332 = 22.332\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 25°C.
\[T(0) = 25 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.29 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 14°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.29 T+14\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.29T+14)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 25 | 0.1 | 0.675 |
| 0.1 | 25.675 | 0.1 | 0.655425 |
| 0.2 | 26.330425 | 0.1 | 0.6364177 |
| 0.3 | 26.9668427 | 0.1 | 0.6179616 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 25+0.675 = 25.675\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 18°C.
\[T(0) = 18 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.28 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 9°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.28 T+9\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.28T+9)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 18 | 0.1 | 0.396 |
| 0.1 | 18.396 | 0.1 | 0.384912 |
| 0.2 | 18.780912 | 0.1 | 0.3741345 |
| 0.3 | 19.1550465 | 0.1 | 0.3636587 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 18+0.396 = 18.396\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 26°C.
\[T(0) = 26 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.49 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.49 T+15\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.49T+15)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 26 | 0.1 | 0.226 |
| 0.1 | 26.226 | 0.1 | 0.214926 |
| 0.2 | 26.440926 | 0.1 | 0.2043946 |
| 0.3 | 26.6453206 | 0.1 | 0.1943793 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 26+0.226 = 26.226\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 23°C.
\[T(0) = 23 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.26 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 12°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.26 T+12\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.26T+12)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 23 | 0.1 | 0.602 |
| 0.1 | 23.602 | 0.1 | 0.586348 |
| 0.2 | 24.188348 | 0.1 | 0.571103 |
| 0.3 | 24.759451 | 0.1 | 0.5562543 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 23+0.602 = 23.602\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 28°C.
\[T(0) = 28 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.3 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 16°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.3 T+16\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.3T+16)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 28 | 0.1 | 0.76 |
| 0.1 | 28.76 | 0.1 | 0.7372 |
| 0.2 | 29.4972 | 0.1 | 0.715084 |
| 0.3 | 30.212284 | 0.1 | 0.6936315 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 28+0.76 = 28.76\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 18°C.
\[T(0) = 18 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.14 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 7°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.14 T+7\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.14T+7)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 18 | 0.1 | 0.448 |
| 0.1 | 18.448 | 0.1 | 0.441728 |
| 0.2 | 18.889728 | 0.1 | 0.4355438 |
| 0.3 | 19.3252718 | 0.1 | 0.4294462 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 18+0.448 = 18.448\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 13°C.
\[T(0) = 13 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.43 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 8°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.43 T+8\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.43T+8)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 13 | 0.1 | 0.241 |
| 0.1 | 13.241 | 0.1 | 0.230637 |
| 0.2 | 13.471637 | 0.1 | 0.2207196 |
| 0.3 | 13.6923566 | 0.1 | 0.2112287 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 13+0.241 = 13.241\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 29°C.
\[T(0) = 29 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.37 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.37 T+15\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.37T+15)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 29 | 0.1 | 0.427 |
| 0.1 | 29.427 | 0.1 | 0.411201 |
| 0.2 | 29.838201 | 0.1 | 0.3959866 |
| 0.3 | 30.2341876 | 0.1 | 0.3813351 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 29+0.427 = 29.427\).
Let’s learn how to use Euler’s method to simulate the changing temperature of a pot of water that is simultaneously cooled by air and warmed by a heater (and well-stirred). The basic idea of Euler’s method is that during a short amount of time, we can assume the average rate of change is approximately equal to the instantaneous rate of change at the beginning of the time interval.
Let’s say that when \(t=0\) min, the water temperature is 23°C.
\[T(0) = 23 \] The air-cooling rate is proportional to the difference between the temperatures; let’s say the proportionality constant for air cooling is 0.48 min\(^{-1}\). Remember, this proportionality constant depends on many things, like the size of pot and speed of wind. Let’s say the heater warms at a steady rate, so if the pot were insulated the temperature would steadily increase at 15°C/min. If we plot temperature versus time, the rate of change is the slope (rise over run), so we know how to calculate the slope \(\frac{\Delta T}{\Delta t}\) if we know the temperature \(T\).
\[\frac{\Delta T}{\Delta t} ~\approx~ -0.48 T+15\]
Let’s use a step size of \(\Delta t=0.1\) min. We typically work in a table (or a spreadsheet). Here are the first few rounds of calculations. Notice it is important to maintain as many digits as possible during the steps.
| \(t\) | \(T\) | \(\Delta t\) | \(\Delta T ~=~ (-0.48T+15)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 23 | 0.1 | 0.396 |
| 0.1 | 23.396 | 0.1 | 0.376992 |
| 0.2 | 23.772992 | 0.1 | 0.3588964 |
| 0.3 | 24.1318884 | 0.1 | 0.3416694 |
Notice, in each row, we add \(T\) and \(\Delta T\) to get the next \(T\). In other words, \(T(t+\Delta t) ~=~ T(t)+\Delta T\). For example \(T(0.1) = 23+0.396 = 23.396\).
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.025v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 700 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 700 | 0.1 | -0.9785941 | -0.0981 |
| 0.2 | -1.9595941 | 699.9019 | 0.1 | -0.9714 | -0.1959594 |
| 0.3 | -2.9309941 | 699.7059406 | 0.1 | -0.9595232 | -0.2930994 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.026v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 500 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 500 | 0.1 | -0.9784979 | -0.0981 |
| 0.2 | -1.9594979 | 499.9019 | 0.1 | -0.971017 | -0.1959498 |
| 0.3 | -2.9305148 | 499.7059502 | 0.1 | -0.9586714 | -0.2930515 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.04v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 700 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 700 | 0.1 | -0.9771506 | -0.0981 |
| 0.2 | -1.9581506 | 699.9019 | 0.1 | -0.9656626 | -0.1958151 |
| 0.3 | -2.9238131 | 699.7060849 | 0.1 | -0.9468053 | -0.2923813 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.027v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 700 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 700 | 0.1 | -0.9784016 | -0.0981 |
| 0.2 | -1.9594016 | 699.9019 | 0.1 | -0.970634 | -0.1959402 |
| 0.3 | -2.9300356 | 699.7059598 | 0.1 | -0.9578202 | -0.2930036 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.048v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 700 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 700 | 0.1 | -0.9763807 | -0.0981 |
| 0.2 | -1.9573807 | 699.9019 | 0.1 | -0.9626096 | -0.1957381 |
| 0.3 | -2.9199902 | 699.7061619 | 0.1 | -0.9400736 | -0.291999 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.043v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 400 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 400 | 0.1 | -0.9768618 | -0.0981 |
| 0.2 | -1.9578618 | 399.9019 | 0.1 | -0.9645171 | -0.1957862 |
| 0.3 | -2.922379 | 399.7061138 | 0.1 | -0.9442767 | -0.2922379 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.033v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 300 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 300 | 0.1 | -0.9778242 | -0.0981 |
| 0.2 | -1.9588242 | 299.9019 | 0.1 | -0.9683379 | -0.1958824 |
| 0.3 | -2.9271621 | 299.7060176 | 0.1 | -0.9527247 | -0.2927162 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.045v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 600 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 600 | 0.1 | -0.9766694 | -0.0981 |
| 0.2 | -1.9576694 | 599.9019 | 0.1 | -0.9637539 | -0.1957669 |
| 0.3 | -2.9214233 | 599.7061331 | 0.1 | -0.9425938 | -0.2921423 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.037v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 300 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 300 | 0.1 | -0.9774393 | -0.0981 |
| 0.2 | -1.9584393 | 299.9019 | 0.1 | -0.9668087 | -0.1958439 |
| 0.3 | -2.925248 | 299.7060561 | 0.1 | -0.9493388 | -0.2925248 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
When a object falls downward, its acceleration is influenced by two competing terms: gravity (\(g\)) and air resistance (\(kv^2\)). Notice air resistance is proportional to the square of velocity.
\[\frac{\Delta v}{\Delta t} = -g+kv^2\]
The coefficient \(k\) depends on the shape of the object, the orientation of the object, the density of the object, and the density of the air.
Because the object is falling straight down, velocity is how fast the height is changing.
\[\frac{\Delta h}{\Delta t} = v\]
Let’s set the parameters and initial values.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(v\) | \(h\) | \(\Delta t\) | \(\Delta v ~=~ (-9.81+0.031v^2)\cdot\Delta t\) | \(\Delta h ~=~ v \cdot \Delta t\) |
|---|---|---|---|---|---|
| 0 | 0 | 500 | 0.1 | -0.981 | 0 |
| 0.1 | -0.981 | 500 | 0.1 | -0.9780167 | -0.0981 |
| 0.2 | -1.9590167 | 499.9019 | 0.1 | -0.969103 | -0.1959017 |
| 0.3 | -2.9281197 | 499.7059983 | 0.1 | -0.954421 | -0.292812 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
The tolerance is \(\pm 0.01\) m.
The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.024\cdot x\cdot (1-x/13)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.5 | 2 | 0.0636923 |
| 2 | 1.5636923 | 2 | 0.066029 |
| 4 | 1.6297214 | 2 | 0.0684199 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.023\cdot x\cdot (1-x/9)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.1 | 2 | 0.0444156 |
| 2 | 1.1444156 | 2 | 0.0459492 |
| 4 | 1.1903647 | 2 | 0.0475145 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.019\cdot x\cdot (1-x/19)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.9 | 2 | 0.06498 |
| 2 | 1.96498 | 2 | 0.0669469 |
| 4 | 2.0319269 | 2 | 0.0689558 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.02\cdot x\cdot (1-x/7)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.1 | 2 | 0.0370857 |
| 2 | 1.1370857 | 2 | 0.0380951 |
| 4 | 1.1751808 | 2 | 0.0391155 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.02\cdot x\cdot (1-x/18)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.7 | 2 | 0.0615778 |
| 2 | 1.7615778 | 2 | 0.0635672 |
| 4 | 1.825145 | 2 | 0.0656032 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.012\cdot x\cdot (1-x/17)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 2.9 | 2 | 0.0577271 |
| 2 | 2.9577271 | 2 | 0.0586351 |
| 4 | 3.0163622 | 2 | 0.0595478 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.012\cdot x\cdot (1-x/15)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 2.6 | 2 | 0.051584 |
| 2 | 2.651584 | 2 | 0.0523886 |
| 4 | 2.7039726 | 2 | 0.053197 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.01\cdot x\cdot (1-x/11)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.8 | 2 | 0.0301091 |
| 2 | 1.8301091 | 2 | 0.0305125 |
| 4 | 1.8606216 | 2 | 0.030918 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.022\cdot x\cdot (1-x/8)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 1.9 | 2 | 0.063745 |
| 2 | 1.963745 | 2 | 0.0651952 |
| 4 | 2.0289402 | 2 | 0.0666321 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The competitive Lotka–Volterra equation is used to simulate population growth.
\[\frac{\Delta x}{\Delta t} = rx\left(1-\frac{x}{K}\right)\]
where \(x\) is population count, \(t\) is time in years, \(r\) is the growth factor, and \(K\) is the carrying capacity.
For our simulation, use the following values for the initial values and parameters.
Using Euler’s method, we can produce the first few rounds of the simulation.
| \(t\) | \(x\) | \(\Delta t\) | \(\Delta x ~=~ 0.015\cdot x\cdot (1-x/8)\cdot\Delta t\) |
|---|---|---|---|
| 0 | 2.5 | 2 | 0.0515625 |
| 2 | 2.5515625 | 2 | 0.0521326 |
| 4 | 2.6036951 | 2 | 0.0526887 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
9 for 9 billion, not \(9\cdot 10^9\) and not \(9000000000\). The tolerance is \(\pm 0.01\) billion.The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.7 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.5c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.6c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.5 b-0.2bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.7 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.8c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.9c\]
Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.4 b-0.2bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.5c\]
Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.3 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.6c\]
Use Euler’s method to estimate the number of bunnies (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand bunnies.
The Lotka–Volterra equations are used to simulate population sizes of predators and prey in an area.
Let’s say prey are bunnies and the predators are coyotes, and the following rate equations hold.
\[\frac{\Delta b}{\Delta t} = 1.6 b-0.3bc\] \[\frac{\Delta c}{\Delta t} = 0.1bc-0.7c\]
Use Euler’s method to estimate the number of coyotes (in thousands) after 10 years. The tolerance is \(\pm 0.01\) thousand coyotes.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = 2.1\] \[y_0 = 0.8\] \[z_0 = -1.1\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.7\] \[y_0 = 1.5\] \[z_0 = -0.2\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.6\] \[y_0 = -2.4\] \[z_0 = 0.6\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.8\] \[y_0 = 2.7\] \[z_0 = -3\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.5\] \[y_0 = 1.1\] \[z_0 = -2.7\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = 0.8\] \[y_0 = 1.6\] \[z_0 = 2.4\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = 1.3\] \[y_0 = 0.1\] \[z_0 = 2\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -0.8\] \[y_0 = -1\] \[z_0 = 1.4\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = -2.3\] \[y_0 = 1.7\] \[z_0 = -0.6\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
In the book “Chaos and Time-Series Analysis” by Julien Clinton Sprott, a simple system of differential equations is given. Research into this system was published by Nosé (1991) and Hoover (1995).
\[\begin{align} \frac{\Delta x}{\Delta t} &= y \\\\ \frac{\Delta y}{\Delta t} &= -x+yz \\\\ \frac{\Delta z}{\Delta t} &= 1-y^2 \end{align}\]
Use these initial conditions: \[t_0 = 0\] \[x_0 = 2\] \[y_0 = -1.1\] \[z_0 = -0.7\]
Use Euler’s method with a step size of \(\Delta t = 0.1\) to simulate until \(t=10\). Plot \(x\), \(y\), and \(z\) as functions of time, using blue, red, and orange respectively. Determine which plot matches your simulation.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.25\sin(4.46z+3.96)+0.48\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:
| \(z\) | \(r=0.25\cdot\sin(4.46z+3.96)+0.48\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.2974854 | 0 | 0.02 | 0.0111209 |
| 0.02 | 0.2829917 | 0.0111209 | 0.02 | 0.0100637 |
| 0.04 | 0.2700646 | 0.0211846 | 0.02 | 0.0091653 |
| 0.06 | 0.2588067 | 0.0303499 | 0.02 | 0.0084171 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.6910107 | ??? | 0.02 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.26\sin(5.61z+1.49)+0.48\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:
| \(z\) | \(r=0.26\cdot\sin(5.61z+1.49)+0.48\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.7391518 | 0 | 0.02 | 0.0686558 |
| 0.02 | 0.7398718 | 0.0686558 | 0.02 | 0.0687896 |
| 0.04 | 0.7373237 | 0.1374454 | 0.02 | 0.0683166 |
| 0.06 | 0.7315397 | 0.205762 | 0.02 | 0.067249 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.669532 | ??? | 0.02 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.3\sin(4.26z+5.12)+0.51\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:
| \(z\) | \(r=0.3\cdot\sin(4.26z+5.12)+0.51\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.2345789 | 0 | 0.02 | 0.0069149 |
| 0.02 | 0.2456981 | 0.0069149 | 0.02 | 0.007586 |
| 0.04 | 0.2587347 | 0.0145009 | 0.02 | 0.0084124 |
| 0.06 | 0.2735942 | 0.0229133 | 0.02 | 0.0094064 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.5234289 | ??? | 0.02 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.12\sin(5.23z+2.82)+0.48\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:
| \(z\) | \(r=0.12\cdot\sin(5.23z+2.82)+0.48\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.5179294 | 0 | 0.02 | 0.0337094 |
| 0.02 | 0.5058353 | 0.0337094 | 0.02 | 0.0321535 |
| 0.04 | 0.4934587 | 0.0658629 | 0.02 | 0.0305993 |
| 0.06 | 0.4809351 | 0.0964622 | 0.02 | 0.0290658 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.597702 | ??? | 0.02 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.16\sin(3.38z+2.85)+0.49\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.05\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.05\). Your table should look like the following:
| \(z\) | \(r=0.16\cdot\sin(3.38z+2.85)+0.49\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.5359965 | 0 | 0.05 | 0.0902555 |
| 0.05 | 0.5095657 | 0.0902555 | 0.05 | 0.0815737 |
| 0.1 | 0.4825775 | 0.1718292 | 0.05 | 0.0731617 |
| 0.15 | 0.4558007 | 0.244991 | 0.05 | 0.0652679 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.4814944 | ??? | 0.05 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.16\sin(5.29z+2.28)+0.49\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.02\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.02\). Your table should look like the following:
| \(z\) | \(r=0.16\cdot\sin(5.29z+2.28)+0.49\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.6114209 | 0 | 0.02 | 0.0469776 |
| 0.02 | 0.5997385 | 0.0469776 | 0.02 | 0.0451995 |
| 0.04 | 0.5868289 | 0.0921771 | 0.02 | 0.0432746 |
| 0.06 | 0.5728364 | 0.1354516 | 0.02 | 0.0412355 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.6435916 | ??? | 0.02 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.36\sin(4.1z+0.11)+0.51\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:
| \(z\) | \(r=0.36\cdot\sin(4.1z+0.11)+0.51\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.5495202 | 0 | 0.04 | 0.075894 |
| 0.04 | 0.6074104 | 0.075894 | 0.04 | 0.0927266 |
| 0.08 | 0.6626865 | 0.1686205 | 0.04 | 0.1103713 |
| 0.12 | 0.7138651 | 0.2789918 | 0.04 | 0.1280773 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.1944835 | ??? | 0.04 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.21\sin(3.9z+0.17)+0.54\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.05\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.05\). Your table should look like the following:
| \(z\) | \(r=0.21\cdot\sin(3.9z+0.17)+0.54\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.5755283 | 0 | 0.05 | 0.1040599 |
| 0.05 | 0.6149594 | 0.1040599 | 0.05 | 0.1188072 |
| 0.1 | 0.6515491 | 0.222867 | 0.05 | 0.1333657 |
| 0.15 | 0.6839106 | 0.3562327 | 0.05 | 0.1469429 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.37186 | ??? | 0.05 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.18\sin(6.93z+5.39)+0.5\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:
| \(z\) | \(r=0.18\cdot\sin(6.93z+5.39)+0.5\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.3597669 | 0 | 0.04 | 0.0325299 |
| 0.04 | 0.3960027 | 0.0325299 | 0.04 | 0.0394127 |
| 0.08 | 0.4401785 | 0.0719426 | 0.04 | 0.0486965 |
| 0.12 | 0.4889217 | 0.120639 | 0.04 | 0.0600784 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.4561006 | ??? | 0.04 |
The tolerance is \(\pm 0.01\) cubic meters.
A potter has thrown a vase with the following profile on the inside. You will determine the volume of water it can hold. Its height is 1 meter tall.
\[r ~=~ 0.13\sin(7.22z+0.62)+0.5\]
To determine the volume, we first break the volume into many discs. In this case, we will estimate the volume using a height of \(\Delta z = 0.04\) for each disc.
Each disc’s volume adds a small amount to the total volume. We know the formula for how much volume is added with each disc.
\[\Delta V = \pi r^2 \cdot \Delta z\]
Estimate the volume of water (in cubic meters) by using \(\Delta z = 0.04\). Your table should look like the following:
| \(z\) | \(r=0.13\cdot\sin(7.22z+0.62)+0.5\) | \(V\) | \(\Delta z\) | \(\Delta V = 2\pi r^2\cdot\Delta z\) |
|---|---|---|---|---|
| 0 | 0.5755346 | 0 | 0.04 | 0.0832497 |
| 0.04 | 0.6025397 | 0.0832497 | 0.04 | 0.0912454 |
| 0.08 | 0.6210517 | 0.1744951 | 0.04 | 0.0969383 |
| 0.12 | 0.6295373 | 0.2714334 | 0.04 | 0.0996054 |
| \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) | \(\vdots\) |
| 1 | 0.6299873 | ??? | 0.04 |
The tolerance is \(\pm 0.01\) cubic meters.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = 0.1\] \[y_0 = -7.5\] \[z_0 = 28.2\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = 14.5\] \[y_0 = -2\] \[z_0 = 26.7\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = 1.8\] \[y_0 = -9.3\] \[z_0 = 29.1\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = -4.2\] \[y_0 = -1.6\] \[z_0 = 32.8\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = -11.6\] \[y_0 = -7.7\] \[z_0 = 23.2\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = 0.6\] \[y_0 = 7.6\] \[z_0 = 33.9\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = -10.6\] \[y_0 = -0.5\] \[z_0 = 35.1\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = -7.5\] \[y_0 = -9.6\] \[z_0 = 35.9\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = 4.4\] \[y_0 = 7\] \[z_0 = 37.1\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Graph \(x\), \(y\), and \(z\) as functions of time, using the initial conditions below:
\[x_0 = -11.2\] \[y_0 = 5.1\] \[z_0 = 24.9\]
Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = -10.4\] \[y_0 = -4.2\] \[z_0 = 38.6\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = -4.7\] \[y_0 = -5.6\] \[z_0 = 29.4\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = -0.5\] \[y_0 = 5.6\] \[z_0 = 26.5\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = -9.7\] \[y_0 = 1.7\] \[z_0 = 24.1\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = 2.1\] \[y_0 = 5.7\] \[z_0 = 24.6\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = 10.3\] \[y_0 = 0.8\] \[z_0 = 24.3\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = 11.5\] \[y_0 = 5.3\] \[z_0 = 16.1\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = 10.1\] \[y_0 = -0.5\] \[z_0 = 21.4\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = -11.9\] \[y_0 = -0.4\] \[z_0 = 24.8\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
By simulating the Lorenz system, Lorenz made an important contribution to the development of chaos theory. Below are the rate equations with his original parameters.
\[\begin{align} \frac{\Delta x}{\Delta t} ~&=~ 10(y-x)\\\\ \frac{\Delta y}{\Delta t} ~&=~ x(28-z)-y\\\\ \frac{\Delta z}{\Delta t} ~&=~ xy-\frac{8}{3}z \end{align}\]
Use Euler’s method with \(\Delta t=0.02\) for time \(t=0\) to \(t=10\). Use the initial conditions below.
\[x_0 = 8.4\] \[y_0 = -0.9\] \[z_0 = 34.2\] Plot \(z\) versus \(x\). (Put \(z\) on the vertical axis and \(x\) on the horizontal axis.) Determine which graph best matches your graph.
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The half-life of an exponentially decaying function is the amount of time it takes for the population to halve.
What is the half-life (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
The doubling time of an exponentially growing function is the amount of time it takes for the population to double.
What is the doubling time (in minutes) of the exponential function graphed above? (The answer is an integer.)
At \(t=0\) a population is \(y(0)=7.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.11 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 7.5e^{0.11t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=6.9\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.09 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 6.9e^{0.09t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=4.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.3 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 4.7e^{0.3t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=7.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.1 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 7.7e^{0.1t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=7.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.21 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 7.5e^{0.21t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=6.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.11 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 6.7e^{0.11t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=12.5\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.15 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 12.5e^{0.15t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=15.9\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.09 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 15.9e^{0.09t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=14.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.12 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 14.7e^{0.12t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=16.7\) million. The population grows with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx 0.26 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 16.7e^{0.26t}\]
What is the doubling time (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=5.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.12 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 5.9e^{-0.12t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=11.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.18 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 11.9e^{-0.18t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=8.2\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.17 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 8.2e^{-0.17t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=15.1\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.18 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 15.1e^{-0.18t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=7.8\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.25 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 7.8e^{-0.25t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=10.9\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.09 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 10.9e^{-0.09t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=12.5\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.14 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 12.5e^{-0.14t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=6\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.33 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 6e^{-0.33t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=6.5\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.09 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 6.5e^{-0.09t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
At \(t=0\) a population is \(y(0)=18.8\) million. The population changes with the following rate equation (assuming \(\Delta t\) is very small): \[\frac{\Delta y}{\Delta t} \approx -0.16 y\]
where \(t\) has units of years, \(y\) has units of millions, and \(k\) has units of years\(^{-1}\). The exact solution (using algebraic calculus instead of Euler’s method) is an exponential function.
\[y = 18.8e^{-0.16t}\]
What is the half life (in years)? (The tolerance is \(\pm 0.2\) years.)
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(6\):
\[\begin{align} f(x) &= 1.81^{x}\\\\ g(x) &= 6\cdot 1.81^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.81^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):
\[\begin{align} f(x) &= 1.78^{x}\\\\ g(x) &= 4\cdot 1.78^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.78^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(6\):
\[\begin{align} f(x) &= 1.87^{x}\\\\ g(x) &= 6\cdot 1.87^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.87^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):
\[\begin{align} f(x) &= 1.5^{x}\\\\ g(x) &= 4\cdot 1.5^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.5^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(3\):
\[\begin{align} f(x) &= 1.22^{x}\\\\ g(x) &= 3\cdot 1.22^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.22^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(3\):
\[\begin{align} f(x) &= 1.31^{x}\\\\ g(x) &= 3\cdot 1.31^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.31^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):
\[\begin{align} f(x) &= 1.32^{x}\\\\ g(x) &= 5\cdot 1.32^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.32^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):
\[\begin{align} f(x) &= 1.27^{x}\\\\ g(x) &= 5\cdot 1.27^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.27^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(5\):
\[\begin{align} f(x) &= 1.48^{x}\\\\ g(x) &= 5\cdot 1.48^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.48^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal shift is equivalent to a vertical stretch. Consider the two functions below, where \(g\) is equivalent to \(f\) stretched vertically by a factor \(4\):
\[\begin{align} f(x) &= 1.41^{x}\\\\ g(x) &= 4\cdot 1.41^{x} \end{align}\]
Instead of using the vertical stretch transformation, we could have used a horizontal shift. Notice the \(g\) curve is congruent to the \(f\) curve, just shifted to the left. Thus, we can express \(g\) as a horizontally shifted version of \(f\).
\[g(x) = f(x+h)\]
Or,
\[g(x) = 1.41^{x+h}\]
where \(h\) represents the distance of the horizontal shift.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(3.1\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/3.1} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.5\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.5} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.5\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.5} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-3.4\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/3.4} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(4.3\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/4.3} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.1\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.1} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(-4.2\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{-x/4.2} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(3.1\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/3.1} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(2.1\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/2.1} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.
For exponential functions, a horizontal stretch is equivalent to a change of base. Consider the two functions below, where \(g\) is equivalent to stretching \(f\) horizontally by a factor \(4.4\):
\[\begin{align} f(x) &= e^{x}\\\\ g(x) &= e^{x/4.4} \end{align}\]
Instead of using the horizontal stretch transformation, we could have changed the base.
\[g(x) = B^{x}\]
where \(B\) represents the new base.